% error in the measurement of
radius of a spere is 2% .
what will be the percentage error
in calculating the volume and
surface area of the spere
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- volume of a sphere =4/3*πr³
- so ,the error in volume =3×Δr/r =3×2%=6%
- surface area =4πr²
- error in surface area is=2Δr/r=2×2=4
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